Question

How do I evaluate `intsqrt(4x^2-9)/xdx`?

Answer 1

The answer is `=sqrt(4x^2-9)-3arctan(1/3sqrt(4x^2-9))+C`

Explanation:

Let `u=sqrt(4x^2-9)`, then

`du=(8xdx)/(2sqrt(4x^2-9))=(4xdx)/sqrt(4x^2-9)`

The integral is

`I=int(sqrt(4x^2-9)dx)/(x)`

`=int(dusqrt(4x^2-9)*sqrt(4x^2-x))/(4x*x)`

`=int(u^2du)/(u^2+9)`

`=int((u^2+9)du)/(u^2+9)-int(9du)/(u^2+9)`

`=intdu-9int(du)/(u^2+9)`

`=u-9int(du)/(u^2+9)`

`I_1-I_2`

The second integral is

`I_2=9int(du)/(u^2+9)`

Let `v=u/3`, `=>`, `dv=(du)/3`

`I_2=9int(3dv)/(9v^2+9)`

`=27/9int(dv)/(v^2+1)`

`=3arctan(v)`

`=3arctan(u/3)`

Finally the integral is

`I=u-3arctan(u/3)`

`=sqrt(4x^2-9)-3arctan(1/3sqrt(4x^2-9))+C`